The Direct Stiffness Method: Temperature Changes & Fabrication Errors in Truss Analysis - 1 | Structural Analysis - Civil Engineering (CE) PDF Download

 Page 1


                                                         
Instructional Objectives 
After reading this chapter the student will be able to 
1. Compute stresses developed in the truss members due to temperature 
changes. 
2. Compute stresses developed in truss members due to fabrication members. 
3. Compute reactions in plane truss due to temperature changes and fabrication 
errors. 
 
 
26.1 Introduction 
In the last four lessons, the direct stiffness method as applied to the truss 
analysis was discussed. Assembly of member stiffness matrices, imposition of 
boundary conditions, and the problem of inclined supports were discussed. Due 
to the change in temperature the truss members either expand or shrink. 
However, in the case of statically indeterminate trusses, the length of the 
members is prevented from either expansion or contraction. Thus, the stresses 
are developed in the members due to changes in temperature. Similarly the error 
in fabricating truss members also produces additional stresses in the trusses. 
Both these effects can be easily accounted for in the stiffness analysis. 
 
 
26.2 Temperature Effects and Fabrication Errors 
 
Page 2


                                                         
Instructional Objectives 
After reading this chapter the student will be able to 
1. Compute stresses developed in the truss members due to temperature 
changes. 
2. Compute stresses developed in truss members due to fabrication members. 
3. Compute reactions in plane truss due to temperature changes and fabrication 
errors. 
 
 
26.1 Introduction 
In the last four lessons, the direct stiffness method as applied to the truss 
analysis was discussed. Assembly of member stiffness matrices, imposition of 
boundary conditions, and the problem of inclined supports were discussed. Due 
to the change in temperature the truss members either expand or shrink. 
However, in the case of statically indeterminate trusses, the length of the 
members is prevented from either expansion or contraction. Thus, the stresses 
are developed in the members due to changes in temperature. Similarly the error 
in fabricating truss members also produces additional stresses in the trusses. 
Both these effects can be easily accounted for in the stiffness analysis. 
 
 
26.2 Temperature Effects and Fabrication Errors 
 
                                                         
 
Consider truss member of length L, area of cross section A as shown in 
Fig.26.1.The change in length l ? is given by 
 
T L l ? = ? a      (26.1) 
 
where a is the coefficient of thermal expansion of the material considered. If the 
member is not allowed to change its length (as in the case of statically 
indeterminate truss) the change in temperature will induce additional forces in the 
member. As the truss element is a one dimensional element in the local 
coordinate system, the thermal load can be easily calculated in global co-
ordinate system by 
 
1
()
t
p AE L ' =?
              (26.2a) 
 
2
()
t
p AE L ' =- ?
              (26.2b) 
 
or 
 
() {}
?
?
?
?
?
?
-
+
? =
1
1
'
L AE p
t               (26.3) 
 
The equation (26.3) can also be used to calculate forces developed in the truss 
member in the local coordinate system due to fabrication error. L ? will be 
considered positive if the member is too long. The forces in the local coordinate 
system can be transformed to global coordinate system by using the equation, 
 
()
()
()
()
()
() ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
t
t
t
t
t
t
p
p
p
p
p
p
'
2
'
1
4
3
2
1
sin 0
cos 0
0 sin
0 cos
?
?
?
?
             (26.4a) 
 
where () ( )
t t
p p
2 1
, and () ( )
t t
p p
4 3
, are the forces in the global coordinate system at 
nodes 1 and 2 of the truss member respectively Using equation (26.3), the 
equation (26.4a) may be written as, 
 
1
2
3
4
() cos
() sin
() cos
() sin
t
t
t
t
p
p
AE L
p
p
?
?
?
?
?? ??
?? ??
?? ? ?
=?
?? ? ?
-
?? ? ?
?? ? ?
-
?? ??
             (26.4b) 
Page 3


                                                         
Instructional Objectives 
After reading this chapter the student will be able to 
1. Compute stresses developed in the truss members due to temperature 
changes. 
2. Compute stresses developed in truss members due to fabrication members. 
3. Compute reactions in plane truss due to temperature changes and fabrication 
errors. 
 
 
26.1 Introduction 
In the last four lessons, the direct stiffness method as applied to the truss 
analysis was discussed. Assembly of member stiffness matrices, imposition of 
boundary conditions, and the problem of inclined supports were discussed. Due 
to the change in temperature the truss members either expand or shrink. 
However, in the case of statically indeterminate trusses, the length of the 
members is prevented from either expansion or contraction. Thus, the stresses 
are developed in the members due to changes in temperature. Similarly the error 
in fabricating truss members also produces additional stresses in the trusses. 
Both these effects can be easily accounted for in the stiffness analysis. 
 
 
26.2 Temperature Effects and Fabrication Errors 
 
                                                         
 
Consider truss member of length L, area of cross section A as shown in 
Fig.26.1.The change in length l ? is given by 
 
T L l ? = ? a      (26.1) 
 
where a is the coefficient of thermal expansion of the material considered. If the 
member is not allowed to change its length (as in the case of statically 
indeterminate truss) the change in temperature will induce additional forces in the 
member. As the truss element is a one dimensional element in the local 
coordinate system, the thermal load can be easily calculated in global co-
ordinate system by 
 
1
()
t
p AE L ' =?
              (26.2a) 
 
2
()
t
p AE L ' =- ?
              (26.2b) 
 
or 
 
() {}
?
?
?
?
?
?
-
+
? =
1
1
'
L AE p
t               (26.3) 
 
The equation (26.3) can also be used to calculate forces developed in the truss 
member in the local coordinate system due to fabrication error. L ? will be 
considered positive if the member is too long. The forces in the local coordinate 
system can be transformed to global coordinate system by using the equation, 
 
()
()
()
()
()
() ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
t
t
t
t
t
t
p
p
p
p
p
p
'
2
'
1
4
3
2
1
sin 0
cos 0
0 sin
0 cos
?
?
?
?
             (26.4a) 
 
where () ( )
t t
p p
2 1
, and () ( )
t t
p p
4 3
, are the forces in the global coordinate system at 
nodes 1 and 2 of the truss member respectively Using equation (26.3), the 
equation (26.4a) may be written as, 
 
1
2
3
4
() cos
() sin
() cos
() sin
t
t
t
t
p
p
AE L
p
p
?
?
?
?
?? ??
?? ??
?? ? ?
=?
?? ? ?
-
?? ? ?
?? ? ?
-
?? ??
             (26.4b) 
                                                         
 
The force displacement equation for the entire truss may be written as, 
 
{} []{} { }
t
p u k p ) ( + =
               (26.5) 
 
where ,
{} p
is the vector of external joint loads applied on the truss and () { }
t
p is the 
vector of joint loads developed in the truss due to change in 
temperature/fabrication error of one or more members. As pointed out earlier. in 
the truss analysis, some joint displacements are known due to boundary 
conditions and some joint loads are known as they are applied 
externally.Thus,one could partition the above equation as, 
 
[] [ ]
[] [ ]
{ }
{}
( )
() ?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
t u
t
k
k
u
u
k
p
p
u
u
k k
k k
p
p
22 21
12 11
    (26.6) 
 
where subscript u is used to denote unknown quantities and subscript k is used 
to denote known quantities of forces and displacements. Expanding equation 
(26.6), 
 
{} []{} [ ] { } ( ) { }
t
k k u k
p u k u k p + + =
12 11
   (26.7a) 
{} []{} [ ] { } ( ) { }
21 22 uu k u
t
pk u k u p =+ +
  (26.7b) 
 
If the known displacement vector 
{ } { } 0 =
k
u
 then using equation (26.2a) the 
unknown displacements can be calculated as 
 
{} [] {} ( ) { } ()
t
k k u
p p k u - =
-1
11
    (26.8a) 
If 
{} 0 ?
k
u
then 
{} [] {} [ ] { } ( ) { } ()
t
k k k u u
p u k p k u - - =
-
12
1
   (26.8b) 
 
After evaluating unknown displacements, the unknown force vectors are 
calculated  using equation (26.7b).After evaluating displacements, the member 
forces in the local coordinate system for each member are evaluated by, 
 
{} [][ ]{} { }
t
p u T k p ' + ' = '
     (26.9a) 
or 
()
() ?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
t
t
p
p
v
u
v
u
L
AE
p
p
'
2
'
1
2
2
1
1
'
2
'
1
sin cos 0 0
0 0 sin cos
1 1
1 1
? ?
? ?
 
Page 4


                                                         
Instructional Objectives 
After reading this chapter the student will be able to 
1. Compute stresses developed in the truss members due to temperature 
changes. 
2. Compute stresses developed in truss members due to fabrication members. 
3. Compute reactions in plane truss due to temperature changes and fabrication 
errors. 
 
 
26.1 Introduction 
In the last four lessons, the direct stiffness method as applied to the truss 
analysis was discussed. Assembly of member stiffness matrices, imposition of 
boundary conditions, and the problem of inclined supports were discussed. Due 
to the change in temperature the truss members either expand or shrink. 
However, in the case of statically indeterminate trusses, the length of the 
members is prevented from either expansion or contraction. Thus, the stresses 
are developed in the members due to changes in temperature. Similarly the error 
in fabricating truss members also produces additional stresses in the trusses. 
Both these effects can be easily accounted for in the stiffness analysis. 
 
 
26.2 Temperature Effects and Fabrication Errors 
 
                                                         
 
Consider truss member of length L, area of cross section A as shown in 
Fig.26.1.The change in length l ? is given by 
 
T L l ? = ? a      (26.1) 
 
where a is the coefficient of thermal expansion of the material considered. If the 
member is not allowed to change its length (as in the case of statically 
indeterminate truss) the change in temperature will induce additional forces in the 
member. As the truss element is a one dimensional element in the local 
coordinate system, the thermal load can be easily calculated in global co-
ordinate system by 
 
1
()
t
p AE L ' =?
              (26.2a) 
 
2
()
t
p AE L ' =- ?
              (26.2b) 
 
or 
 
() {}
?
?
?
?
?
?
-
+
? =
1
1
'
L AE p
t               (26.3) 
 
The equation (26.3) can also be used to calculate forces developed in the truss 
member in the local coordinate system due to fabrication error. L ? will be 
considered positive if the member is too long. The forces in the local coordinate 
system can be transformed to global coordinate system by using the equation, 
 
()
()
()
()
()
() ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
t
t
t
t
t
t
p
p
p
p
p
p
'
2
'
1
4
3
2
1
sin 0
cos 0
0 sin
0 cos
?
?
?
?
             (26.4a) 
 
where () ( )
t t
p p
2 1
, and () ( )
t t
p p
4 3
, are the forces in the global coordinate system at 
nodes 1 and 2 of the truss member respectively Using equation (26.3), the 
equation (26.4a) may be written as, 
 
1
2
3
4
() cos
() sin
() cos
() sin
t
t
t
t
p
p
AE L
p
p
?
?
?
?
?? ??
?? ??
?? ? ?
=?
?? ? ?
-
?? ? ?
?? ? ?
-
?? ??
             (26.4b) 
                                                         
 
The force displacement equation for the entire truss may be written as, 
 
{} []{} { }
t
p u k p ) ( + =
               (26.5) 
 
where ,
{} p
is the vector of external joint loads applied on the truss and () { }
t
p is the 
vector of joint loads developed in the truss due to change in 
temperature/fabrication error of one or more members. As pointed out earlier. in 
the truss analysis, some joint displacements are known due to boundary 
conditions and some joint loads are known as they are applied 
externally.Thus,one could partition the above equation as, 
 
[] [ ]
[] [ ]
{ }
{}
( )
() ?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
t u
t
k
k
u
u
k
p
p
u
u
k k
k k
p
p
22 21
12 11
    (26.6) 
 
where subscript u is used to denote unknown quantities and subscript k is used 
to denote known quantities of forces and displacements. Expanding equation 
(26.6), 
 
{} []{} [ ] { } ( ) { }
t
k k u k
p u k u k p + + =
12 11
   (26.7a) 
{} []{} [ ] { } ( ) { }
21 22 uu k u
t
pk u k u p =+ +
  (26.7b) 
 
If the known displacement vector 
{ } { } 0 =
k
u
 then using equation (26.2a) the 
unknown displacements can be calculated as 
 
{} [] {} ( ) { } ()
t
k k u
p p k u - =
-1
11
    (26.8a) 
If 
{} 0 ?
k
u
then 
{} [] {} [ ] { } ( ) { } ()
t
k k k u u
p u k p k u - - =
-
12
1
   (26.8b) 
 
After evaluating unknown displacements, the unknown force vectors are 
calculated  using equation (26.7b).After evaluating displacements, the member 
forces in the local coordinate system for each member are evaluated by, 
 
{} [][ ]{} { }
t
p u T k p ' + ' = '
     (26.9a) 
or 
()
() ?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
t
t
p
p
v
u
v
u
L
AE
p
p
'
2
'
1
2
2
1
1
'
2
'
1
sin cos 0 0
0 0 sin cos
1 1
1 1
? ?
? ?
 
                                                         
Expanding the above equation, yields 
 
{}{} L AE
v
u
v
u
L
AE
p ? +
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - = '
2
2
1
1
1
sin cos sin cos ? ? ? ?
   (26.10a) 
And, 
 
{}{}
1
1
2
2
2
cos sin cos sin
u
v
AE
pAEL
u L
v
?? ? ?
??
??
??
'=- - - ?
??
??
??
??
   (26.10b) 
Few problems are solved to illustrate the application of the above procedure to 
calculate thermal effects /fabrication errors in the truss analysis:- 
 
Example 26.1 
Analyze the truss shown in Fig.26.2a, if the temperature of the member (2) is 
raised by C
o
40 .The sectional areas of members in square centimeters are 
shown in the figure. Assume 
2 5
/ 10 2 mm N E × = and 
1/ 75,000 a =
 per C
o
. 
 
 
Page 5


                                                         
Instructional Objectives 
After reading this chapter the student will be able to 
1. Compute stresses developed in the truss members due to temperature 
changes. 
2. Compute stresses developed in truss members due to fabrication members. 
3. Compute reactions in plane truss due to temperature changes and fabrication 
errors. 
 
 
26.1 Introduction 
In the last four lessons, the direct stiffness method as applied to the truss 
analysis was discussed. Assembly of member stiffness matrices, imposition of 
boundary conditions, and the problem of inclined supports were discussed. Due 
to the change in temperature the truss members either expand or shrink. 
However, in the case of statically indeterminate trusses, the length of the 
members is prevented from either expansion or contraction. Thus, the stresses 
are developed in the members due to changes in temperature. Similarly the error 
in fabricating truss members also produces additional stresses in the trusses. 
Both these effects can be easily accounted for in the stiffness analysis. 
 
 
26.2 Temperature Effects and Fabrication Errors 
 
                                                         
 
Consider truss member of length L, area of cross section A as shown in 
Fig.26.1.The change in length l ? is given by 
 
T L l ? = ? a      (26.1) 
 
where a is the coefficient of thermal expansion of the material considered. If the 
member is not allowed to change its length (as in the case of statically 
indeterminate truss) the change in temperature will induce additional forces in the 
member. As the truss element is a one dimensional element in the local 
coordinate system, the thermal load can be easily calculated in global co-
ordinate system by 
 
1
()
t
p AE L ' =?
              (26.2a) 
 
2
()
t
p AE L ' =- ?
              (26.2b) 
 
or 
 
() {}
?
?
?
?
?
?
-
+
? =
1
1
'
L AE p
t               (26.3) 
 
The equation (26.3) can also be used to calculate forces developed in the truss 
member in the local coordinate system due to fabrication error. L ? will be 
considered positive if the member is too long. The forces in the local coordinate 
system can be transformed to global coordinate system by using the equation, 
 
()
()
()
()
()
() ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
?
?
?
?
t
t
t
t
t
t
p
p
p
p
p
p
'
2
'
1
4
3
2
1
sin 0
cos 0
0 sin
0 cos
?
?
?
?
             (26.4a) 
 
where () ( )
t t
p p
2 1
, and () ( )
t t
p p
4 3
, are the forces in the global coordinate system at 
nodes 1 and 2 of the truss member respectively Using equation (26.3), the 
equation (26.4a) may be written as, 
 
1
2
3
4
() cos
() sin
() cos
() sin
t
t
t
t
p
p
AE L
p
p
?
?
?
?
?? ??
?? ??
?? ? ?
=?
?? ? ?
-
?? ? ?
?? ? ?
-
?? ??
             (26.4b) 
                                                         
 
The force displacement equation for the entire truss may be written as, 
 
{} []{} { }
t
p u k p ) ( + =
               (26.5) 
 
where ,
{} p
is the vector of external joint loads applied on the truss and () { }
t
p is the 
vector of joint loads developed in the truss due to change in 
temperature/fabrication error of one or more members. As pointed out earlier. in 
the truss analysis, some joint displacements are known due to boundary 
conditions and some joint loads are known as they are applied 
externally.Thus,one could partition the above equation as, 
 
[] [ ]
[] [ ]
{ }
{}
( )
() ?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
t u
t
k
k
u
u
k
p
p
u
u
k k
k k
p
p
22 21
12 11
    (26.6) 
 
where subscript u is used to denote unknown quantities and subscript k is used 
to denote known quantities of forces and displacements. Expanding equation 
(26.6), 
 
{} []{} [ ] { } ( ) { }
t
k k u k
p u k u k p + + =
12 11
   (26.7a) 
{} []{} [ ] { } ( ) { }
21 22 uu k u
t
pk u k u p =+ +
  (26.7b) 
 
If the known displacement vector 
{ } { } 0 =
k
u
 then using equation (26.2a) the 
unknown displacements can be calculated as 
 
{} [] {} ( ) { } ()
t
k k u
p p k u - =
-1
11
    (26.8a) 
If 
{} 0 ?
k
u
then 
{} [] {} [ ] { } ( ) { } ()
t
k k k u u
p u k p k u - - =
-
12
1
   (26.8b) 
 
After evaluating unknown displacements, the unknown force vectors are 
calculated  using equation (26.7b).After evaluating displacements, the member 
forces in the local coordinate system for each member are evaluated by, 
 
{} [][ ]{} { }
t
p u T k p ' + ' = '
     (26.9a) 
or 
()
() ?
?
?
?
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
=
?
?
?
?
?
?
?
?
?
?
t
t
p
p
v
u
v
u
L
AE
p
p
'
2
'
1
2
2
1
1
'
2
'
1
sin cos 0 0
0 0 sin cos
1 1
1 1
? ?
? ?
 
                                                         
Expanding the above equation, yields 
 
{}{} L AE
v
u
v
u
L
AE
p ? +
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - = '
2
2
1
1
1
sin cos sin cos ? ? ? ?
   (26.10a) 
And, 
 
{}{}
1
1
2
2
2
cos sin cos sin
u
v
AE
pAEL
u L
v
?? ? ?
??
??
??
'=- - - ?
??
??
??
??
   (26.10b) 
Few problems are solved to illustrate the application of the above procedure to 
calculate thermal effects /fabrication errors in the truss analysis:- 
 
Example 26.1 
Analyze the truss shown in Fig.26.2a, if the temperature of the member (2) is 
raised by C
o
40 .The sectional areas of members in square centimeters are 
shown in the figure. Assume 
2 5
/ 10 2 mm N E × = and 
1/ 75,000 a =
 per C
o
. 
 
 
                                                         
 
The numbering of joints and members are shown in Fig.26.2b. The possible 
global displacement degrees of freedom are also shown in the figure. Note that 
lower numbers are used to indicate unconstrained degrees of freedom. From the 
figure it is obvious that the displacements 
0
8 7 6 5 4 3
= = = = = = u u u u u u
due to 
boundary conditions. 
The temperature of the member (2) has been raised by C
o
40 . Thus, 
 
T L L ? = ? a 
( )()
3
10 2627 . 2 40 2 3
75000
1
-
× = = ?L
m    (1) 
 
The forces in member (2) due to rise in temperature in global coordinate system 
can be calculated using equation (26.4b).Thus, 
 
()
()
()
?
?
?
?
?
?
?
?
?
?
?
?
?
?
-
-
? =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
sin
cos
sin
cos
) (
2
1
6
5
L AE
p
p
p
p
t
t
t
t
     (2) 
 
For member (2), 
 
2 4 2
10 20 20 m cm A
-
× = = and 
o
45 = ?